Questions: Homework #8 Question 10 of 20 (2 points) I Question Attempt: 1 of Unlimited Graph the system below. Then write its solution. y=x^2 y=-2x+3 If there is more than one solution, use the "or" button. Solution(s):

Solution Steps

The system of equations given is:

1. \( y = x^2 \)
2. \( y = -2x + 3 \)

To find the solution(s), set the equations equal to each other: \[ x^2 = -2x + 3 \]

Rearrange the equation: \[ x^2 + 2x - 3 = 0 \]

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -3 \): \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \] \[ x = \frac{-2 \pm \sqrt{4 + 12}}{2} \] \[ x = \frac{-2 \pm \sqrt{16}}{2} \] \[ x = \frac{-2 \pm 4}{2} \]

The solutions for \( x \) are: \[ x = 1 \quad \text{or} \quad x = -3 \]

For \( x = 1 \): \[ y = 1^2 = 1 \]

For \( x = -3 \): \[ y = (-3)^2 = 9 \]

Final Answer